The line `x=2y` intersects the ellipse `(x^(2))/4+y^(2)=1` at the points `P` and `Q`. The equation of the circle with `PQ` as diameter is

Options

A. `x^(2)+y^(2)=1/2`
B. `x^(2)+y^(2)=1`
C. `x^(2)+y^(2)=2`
D. `x^(2)+y^(2)=5/2`

Answer

D

Explanation

We have `x=2y`……………..i
and `(x^(2))/4+y^(2)=1`………………ii
On putting `x=2y` in eq. ii we get
`((2y)^(2))/4+y^(2)=1implies(4y^(2))/4+y^(2)=1`
`implies2y^(2)=1impliesy=+-1/(sqrt(2))`
From eq. i `x=+-sqrt(2)`
`:.P(sqrt(2),1/(sqrt(2)))`and `Q(-sqrt(2),-1/(sqrt(2)))`
Equation of circle is
`(x-sqrt(2))(x+sqrt(2))+(y-1/(sqrt(2)))`
`(y+1/(sqrt(2)))=0,x^(2)-2+y^(2)-1/2=0`
`impliesx^(2)+y^(2)=5//2`